Falling Body with Air Resistance Falling Body with Air Resistance Assume that a body of mass m is dropped from a great height above the surface of the earth. Our task is to answer the following Analyze the forces of gravity and air resistance. Use Newton's Law force = mass x acceleration to write down an equation that relates vertical speed with vertical acceleration. This will be a simple kind of differential equation . Do some calculus magic to find an explicit formula for the vertical speed ---here we use stuff from differential calculus (Math 251) to solve the differential equation. Introduce some new calculus magic to answer the important question---this material is discussed in Section 5.1 of your text; it involves the introduction of a rectangle approximation technique to estimate the area under the speed graph. Gravity Gravity makes the object fall. It was Galileo who offered the hypothesis that the acceleration due to gravity near the Earth's surface is essentially constant. Newton's law, force = mass x acceleration, then implies that Fgrav = mg where m is the mass of the object kilograms and g is the gravitational acceleration near the earth's surface, which is about 10 m/sec2.

The basic unit of force in the metric system is the newton, which is one kg*m/sec2. Both m and g are taken to be positive. Air Resistance Let s(t) be the downward speed of the object t seconds after it is dropped, measured in m/sec. So we are assuming that s(0) = 0. One hypothesis governing air resistance is the following: "Linear" Air Resistance Hypothesis The force due to air resistance is proportional to the speed, and is applied in the direction opposite to motion. Look at it this way, as the object moves through the air, it collides with air molecules, displacing them as it falls. The faster the object moves, the more collisions and so the greater the overall force due to air resistance. An alternative hypothesis is that the force due to air resistance is proportional to the square of the speed...this might be more reasonable, for example, at very high speeds. But we will stick with the linear assumption. This linear hypothesis means that there is a positive constant k such that Fair = -ks(t) for all t.

The positive constant k is the constant of proportionality and its units are kg/sec; the numerical value of k depends upon the shape of the object being dropped and the density of the atmosphere. Total Force, Acceleration, and Speed We now see that the total force acting on the falling object at any time t is given by Ftot = Fgrav + Fair = mg - ks(t) Now comes the first real conclusion involving calculus. Remember that acceleration is the rate of change of speed. (Actually it's better to think of acceleration as the rate of change of velocity, but we can identify the two concepts in this example since the object only moves in one direction.) Thus, if a(t) is the acceleration on the object at time t, then a(t) = s'(t). Now we use Newton's Law to conclude that Ftot = ma(t), or more precisely, we obtain the following relationship between speed s(t) and acceleration s'(t). Differential Equation for Speed at Time t s(t) = (mg/k)(1 - e(-k/m)t) A Neat Thing to Notice: Terminal Velocity Observe that under our assumptions (the forces acting are gravity and "linear" air resistance) we discover the phenomenon of terminal velocity.

This comes about by asking for the limit of speed at time goes to infinity. As t goes to infinitey, the term e(-k/m)t goes to zero since k and m are positive. So we find terminal velocity = limt-->infty s(t) = mg/k The phenomenon of terminal velocity is thus discovered as a consequence of our force assumptions. (!!!) A Sample Graph of Speed vs Time Using the specific values g = 10 m/sec2, m = 1 kg, and k = 0.1 kg/sec, the formula for the speed function and its graph look like this: 4 Lift, Thrust, Weight, and Drag Copyright © 1996-2005 jsd 4 Lift, Thrust, Weight, and Drag It is better to be on the ground wishing you were flying, rather than up in the air wishing you were on the ground. — Aviation proverb. 4.1 Definitions The main purpose of this chapter is to clarify the concepts of lift, drag, thrust, and weight. Pilot books call these the four forces . It is not necessary for pilots to have a super-precise understanding of the four forces.

The concept of energy (discussed in chapter 1 ) is considerably more important. In the cockpit (especially in critical situations like final approach) I think about the energy budget a lot, and think about forces hardly at all. Still, there are a few situations that can be usefully discussed in terms of forces, so we might as well learn the terminology. The relative wind acting on the airplane produces a certain amount of force which is called (unsurprisingly) the total aerodynamic force . This force can be resolved into components, called lift and drag, as shown in figure 4.1 . Figure 4.1 : Total Aerodynamic Force = Lift + Drag Here are the official, conventional definitions of the so-called four forces: Drag is a component of the aerodynamic force, namely the projection onto the direction parallel to the relative wind. Lift is another component of the aerodynamic force, namely the projection onto the two directions perpendicular to the relative wind. Weight is the force of gravity. It is equal to the mass of the airplane times the local gravitational acceleration, i.e. the local gravitational field. You can measure the acceleration of gravity by observing the motion – relative to your chosen reference frame – of a freely falling object.

This defines what we mean by “downward”. For present purposes, it is convenient to choose a reference frame attached to a nearby point on the earth’s surface, in which case the downward direction points approximately toward the center of the earth. In later chapters we will sometimes find it convenient to choose a frame comoving with the airplane, in which case our notions of “gravity” and and “downward” are heavily modified. That’s because these notions are totally dependent on the choice of reference frame. For details on this, see section 19.6 . Thrust is the force produced by the engine. It is directed forward along the axis of the engine. It is ironic that according to convention, the total aerodynamic force is not listed among the four forces. In general, the lift is a vector in two dimensions, i.e. a vector in the plane perpendicular to the relative wind; this plane is shown in yellow in figure 4.1 . In general, you get to specify the magnitude and direction of this vector. However, in normal coordinated flight, to a good approximation, the lift is perpendicular to the wingspan (as well as perpendicular to the relative wind).

Therefore, when people speak of the lift in one-dimensional terms, they are probably talking about the component of lift in the direction perpendicular to the wingspan (unless context demands otherwise). Figure 4.2 : The Four Forces — Low Speed Descent Figure 4.2 shows the orientation of the four forces when the airplane is in “slow flight”, i.e. descending with a nose-high attitude, with the engine producing some power. Similarly, figure 4.3 shows the four forces when airplane in a high-speed descent. The angle of attack is much lower, which is consistent with the higher airspeed. Finally, figure 4.4 shows the four forces when the airplane is in a climb. I have chosen the angle of attack, the lift, and the drag to have the same magnitude as in figure 4.3 . Figure 4.3 : The Four Forces — High Speed Dive Figure 4.4 : The Four Forces — Climb Note that the four forces are defined with respect to three different coordinate systems: lift and drag are defined relative to the wind, gravity is sometimes defined relative to the earth, and thrust is defined relative to the orientation of the engine. This makes things complicated.

For example, in figure 4.2 you can see that thrust, lift and drag all have vertical components that combine to oppose the weight. Meanwhile the thrust and lift both have forward horizontal components. 4.2 Balance of Forces Let’s temporarily imagine you are flying straight and level, maintaining constant speed and constant attitude, through still air. We further imagine that the axis of the engine happens to be aligned with the straight-ahead direction, for this chosen attitude. Then all three coordinate systems coincide, in which case thrust is opposite to drag, and lift is opposite to weight. In reality, it isn’t safe to assume that lift always matches weight, or thrust exactly matches drag. Consider a bomb falling straight down ( figure 4.5 ) – it has no lift and no thrust; when it reaches terminal velocity its weight is supported purely by drag. Another interesting case is a moon lander hovering on its rocket plume ( figure 4.6 ) — it has no lift and no drag; its weight is supported by its thrust. Figure 4.6 : Moon Lander (Weight = Thrust) You may think lift, thrust, weight, and drag are defined in a crazy way, but the definitions aren’t going to change anytime soon.

They have too much history behind them, and they actually have advantages when analyzing complex situations. The good news is that these subtleties usually don’t bother you. First of all, the angles in figure 4.2 are greatly exaggerated. In ordinary transportation (as opposed to aerobatics), even in climbs and descents, the pitch angle is always rather small, so thrust is always nearly horizontal. Also, the relative wind differs from horizontal by only a few degrees, so drag is always nearly horizontal, and lift is nearly vertical except in turns. If we don’t like the technical definitions of lift, drag, thrust and weight, we are free to use other terms. For example, we can make the following sweeping statement: in unaccelerated flight, the upward forces balance the downward forces, and the forward forces balance the rearward forces. This statement is true whether or not we calculate separately the contributions of lift, drag, thrust and weight. This statement is simple because it expresses all the forces in terms of the horizontal and vertical components, without reference to the direction of flight. Alternatively, we can choose, if we want, express all the forces relative to the direction of flight, with minimal references to horizontal and vertical.

An example of this is shown in figure 4.7 . We resolve the weight into two components: The parallel component of weight is parallel to the direction of flight, while the perpendicular component of weight is at right angles to the direction of flight. These two components added together are equal to the total weight. (The total weight is directly downward, but we don’t care about that in figure 4.7 , because we are using the components instead.) Figure 4.7 : Force Resolved into Components This description is nice because it shows things from the pilot’s point of view, which is consistent with the general spirit of this book. Note that in figure 4.7 the parallel component of weight is pulling the aircraft forward along the path of flight. Indeed, in this situation, this component of weight is a larger contribution than engine thrust. Before going on, let me mention a couple of petty paradoxes. (1) In a steady, low-speed, high-power, nose-up climb, lift is less than weight. That’s possible because thrust is supporting part of the weight. It sounds crazy to say that lift is less than weight during climb, but it is technically true.

(2) In a steady, low-power, high-speed, nose-down descent, lift is once again less than weight. That’s possible because drag is supporting part of the weight. These paradoxes are pure technicalities, consequences of the peculiar definitions of the four forces. They have no impact on pilot technique. There is some additional discussion of the balance of forces in section 19.1 . 4.3 Forces During a Turn The most important non-aerobatic situation where you have to worry about the forces on the airplane is during a turn. In a steeply-banked turn, the lift vector is inclined quite a bit to the left or right of vertical. In order to support the weight of the airplane and pull the airplane around the turn, the lift must be significantly greater than the weight. This leads us to the notion of load factor , which is discussed in section 6.2.3 . The bottom line is that thrust is usually nearly equal (and opposite) to drag, and lift is usually nearly equal (and opposite) to weight times load factor . In a turn, it is sometimes useful to express the total lift as a sum of two components.

The vertical component of lift, as usual, is what opposes weight, so there is no net vertical force, so that the airplane does not accelerate upwards or downward. The horizontal component of lift is what provides a horizontal force that changes which way the airplane is going. 1 In a steeply-banked turn, the horizontal component of lift is quite large. In the pilot’s frame of reference, that means the airplane is subject to very significant centrifugal forces. This important and interesting topic will be discussed in section 6.2 . 4.4 Types of Drag We have seen that the total force on the airplane can be divided into lift and drag. We now explore various ways of subdividing and classifying the drag. When a force acts on a surface, it is often useful to distinguish processes that act perpendicular to the surface (pressure against the surface) versus forces that act parallel to the surface (friction along the surface). Figure 4.10 : Friction Drag Airflow Orientation Friction drag is proportional to viscosity (roughly, the “stickiness” of the fluid). Fortunately, air has a rather low viscosity, so in most situations friction drag is small compared to pressure drag.

In contrast, pressure drag depends on the mass density (not viscosity) of the air. Friction drag and pressure drag both create a force in proportion to the area involved, and to the square of the airspeed. Part of the pressure drag that a wing produces depends on the amount of lift it is producing. This part of the drag is called induced drag . The rest of the drag — everything except induced drag — is called parasite drag . The various categories of drag are summarized in figure 4.11 . Figure 4.11 : Categories of Drag The part of the parasite drag that is not due to friction is called form drag . That is because it is extremely sensitive to the detailed form and shape of airplane, as we now discuss. A non-streamlined object (such as the flat plate in figure 4.12 ) can have ten times more form drag than a streamlined object of comparable frontal area (such as the one shown in figure 4.13 ).

The peak pressure in front of the two shapes will be the same, but (1) the streamlined shape causes the air to accelerate, so the region of highest pressure is smaller, and more importantly, (2) the streamlined shape cultivates high pressure behind the object that pushes it forward, canceling most of the pressure drag, as shown in figure 4.13 . This is called pressure recovery . Figure 4.13 : Pressure Recovery Any object moving through the air will have a high-pressure region in front, but a properly streamlined object will have a high-pressure region in back as well, resulting in pressure recovery. The flow pattern 2 near a non-streamlined object is not symmetric fore-and-aft because the stream lines separate from the object as they go around the sharp corners of the plate. Separation is discussed at more length in chapter 18 . Streamlining is never perfect; there is always at least some net pressure drag. Induced drag also contributes to the pressure drag whenever lift is being produced (even for perfectly streamlined objects in the absence of separation). Except for very small objects and/or very low speeds, pressure drag is larger than friction drag (even for well-streamlined objects).

The pressure drag of a non-streamlined object is much larger still. This is why on high-performance aircraft, people go to so much trouble to ensure that even the smallest things (e.g. fuel-cap handles) are perfectly aligned with the airflow. An important exception involves the air that has to flow through the engine compartment to cool the engine. A lot of air has to flow through narrow channels. The resulting friction drag — called cooling drag — amounts to 30% of the total drag of some airplanes. Unlike pressure drag, friction drag cannot possibly be canceled, even partially. Friction drag causes energy to be thrown overboard. The energy gets carried away by the relative wind, and there is no practical way for the airplane to recover that energy. The way to reduce induced drag (while maintaining the same amount of lift) is to have a longer wingspan and/or to fly faster. The way to minimize friction drag is to minimize the total wetted area (i.e. the total area that has high-speed air flowing along it). The way to reduce form drag is to minimize separation, by making everything streamlined. 4.5 Coefficients, Forces, and Power The word “drag”, by itself, usually refers to a force (the force of drag).

Similarly, the word “lift”, by itself, usually refers to a force. However, there are other ways of looking at things. * Coefficients It is often convenient to write the drag force as a dimensionless number (the coefficient of drag ) times a bunch of factors that characterize the situation: drag force = ½ρV2 × coefficient of drag × area ( 4.1 ) where ρ (the Greek letter “rho”) is the density of the air, V is your true airspeed, and the relevant area is typically taken to be the wing area (excluding the surface area of the fuselage, et cetera). Similarly, there is a coefficient of lift : lift force = ½ρV2 × coefficient of lift × area ( 4.2 ) We used these equations back in section 2.13 to explain why the airspeed indicator is a good source of information about angle of attack. One nice thing about these equations is that the coefficient of lift and the coefficient of drag depend on the angle of attack and not much else.

If you could (by magic) hold the angle of attack constant, the coefficient of lift and the coefficient of drag would be remarkably independent of airspeed, density , temperature , or whatever. The coefficient of lift is a ratio 3 that basically measures how effectively the wing turns the available dynamic pressure into useful average suction over the wing. A typical airfoil can achieve a coefficient of lift around 1.5 without flaps; even with flaps it is hard to achieve a coefficient of lift bigger than 2.5 or so. For data on real airfoils, see figure 3.16 and/or reference 26 . Figure 4.14 shows how the various coefficients depend on angle of attack. The left side of the figure corresponds to the highest airspeeds (lowest angles of attack). Note that the coefficient-of-lift curve has been scaled down by a factor of ten to make it fit on the same graph as the other curves. Airplanes are really good at making lots of lift with little drag. In the range corresponding to normal flight (say 10 degrees angle of attack or less) we can use the basic lift/drag model.

The details of this model are explored in section 7.6.3 , but in most piloting situations all you need to know are the following approximations, which are the conceptual basis of the model: the coefficient of lift is proportional to the angle of attack, the coefficient of induced drag is proportional to the square of the angle of attack, and the coefficient of parasite drag is essentially constant. Figure 4.14 : Coefficients versus Angle of Attack In flight, we are not free to make any amount of lift we want. The lift is nearly always equal to the weight times the load factor. This leads us to rearrange the lift equation as follows: coefficient of lift = (weight × load factor) / (½ρV2 × area) ( 4.3 ) On the right-hand side of this equation, the only factors that are likely to change from moment to moment are airspeed and load factor. (Weight can change, too, but usually only slowly. Extending the flaps can change the area somewhat.) Because of the factor of airspeed squared, the airplane must fly at a very high coefficient of lift in order to support its weight at low airspeeds.

These ideas can be used to explain the shape of the power curve, as shown in figure 4.15 (which is a copy of figure 1.13 ). Figure 4.15 : Power Curve — Three Regimes As you can see from the green curve in figure 4.14 , at high angles of attack, beyond the critical angle of attack, the coefficient of lift is a decreasing function of angle of attack. The lower coefficient of lift means the airspeed must be higher, in order to support the weight of the airplane. This explains why the stalled regime of the power curve lies to the right of the critical point, as you can see in figure 4.15 . Meanwhile, at high angles of attack, the drag is very high, as shown by the red curve in figure 4.14 . This explains why the the stalled regime of the power curve lies below the critical point. For more on this, see section 5.3 . Figure 4.16 is related to figure 4.14 . It plots the same four curves against airspeed (rather than angle of attack). Now the left side of the plot corresponds to the lowest airspeeds (highest angles of attack).

Figure 4.16 : Coefficients versus Airspeed At higher angles of attack (approaching or exceeding the critical angle of attack) the basic-model approximations break down. The coefficient of parasite drag will rapidly become quite large, and the induced drag will probably be quite large also. There will be no simple proportionality relationships. The details aren’t of much interest to most pilots, for the following reason: Typically you recover from a stall as soon as you notice it, so you don’t spend much time in the stalled regime. If you do happen to be interested in stalled flight and spins, see chapter 18 . * Forces Figure 4.17 shows the corresponding forces. We see that whereas the coefficient of parasite drag was more or less constant, the force of parasite drag increases with airspeed. If somebody says “the drag is a ... function of airspeed” you have to ask whether “drag” refers to the drag coefficient, the drag force, or (as discussed below) the drag power.

Figure 4.17 : Forces versus Airspeed We can also see in the figure that the lift force curve is perfectly constant, which is reassuring, since the figure was constructed using the principle that the lift force must equal the weight of the airplane; this is how I converted angle of attack to airspeed. The lowest point in the total drag force curve corresponds to VL/D, and gives the best lift-to-drag ratio. Using the standard lift/drag model and a little calculus, it can be shown that this occurs right at the point where the induced drag force curve crosses the parasite drag force curve. * Powers Figure 4.18 shows the amount of dissipation due to drag, for the various types of drag. Dissipation is a form of power, i.e. energy per unit time. Figure 4.18 : Powers versus Airspeed Dissipation is related to force by the simple rule: power = force · velocity ( 4.4 ) In this equation, we are multiplying two vectors using the dot product (·) , 4 which means that only the velocity component in the direction of the force counts.

In the case of drag, we have specifically: dissipation = force of drag · airspeed ( 4.5 ) The lowest point in the curve for total drag power corresponds to VY, and gives the best rate of climb. Using the standard lift/drag model and a little calculus, it can be shown that at this speed, the minimum occurs right at the point where the induced drag power is 3/4ths of the total, and the parasite drag power is 1/4th of the total. Actually, in the airplane represented in these figures, VY is so close to the stalling speed that the standard lift/drag model is starting to break down, and the 3:1 ratio is not exactly accurate. In the case of lift, the lift force is (by its definition) perpendicular to the relative wind, so there is no such thing as dissipation due to lift. (Of course the physical process that produces lift also produces induced drag, but the part of the force properly called lift isn’t the part that contributes to the power budget.) 4.6 Induced vs. Parasite Drag There are several useful conclusions we can draw from these curves.

For starters, we see that the curve of total power required to overcome dissipation has a familiar shape; it is just an upside-down version of the power curve that appears in section 1.2.5 and elsewhere throughout this book. We can also see why the distinction between induced drag and parasite drag is significant to pilots: In the mushing regime, most of the drag is induced drag. As you go slower and slower, induced drag increases dramatically and parasite drag becomes almost negligible. At high airspeeds, parasite drag is dominant and induced drag becomes almost negligible. In the high-speed regime (which includes normal cruise), the power required increases rapidly with increasing airspeed. Eventually it grows almost like the cube of the airspeed. The reason is easy to see: parasite drag is the dominant contribution to the coefficient of drag in this regime, and is more-or-less independent of airspeed. 5 We pick up two factors of V from equation 4.1 and one from equation 4.4 .

Knowing this cube law is useful for figuring out the shape of your airplane’s power curve ( section 7.6.2 ), and for figuring out how big an engine you need as a function of speed ( section 7.6.4 ) and altitude ( section 7.6.5 ). Free Fall and Air Resistance Newton's Laws - Lesson 3 - Newton's Second Law of Motion Free Fall and Air Resistance Double Trouble In a previous unit, it was stated that all objects ( regardless of their mass ) free fall with the same acceleration - 9.8 m/s/s. This particular acceleration value is so important in physics that it has its own peculiar name - the acceleration of gravity - and its own peculiar symbol - g. But why do all objects free fall at the same rate of acceleration regardless of their mass? Is it because they all weigh the same? ... because they all have the same gravity? ... because the air resistance is the same for each? Why? These questions will be explored in this section of Lesson 3. In addition to an exploration of free fall, the motion of objects that encounter air resistance will also be analyzed. In particular, two questions will be explored: Why do objects that encounter air resistance ultimately reach a terminal velocity?

In situations in which there is air resistance, why do more massive objects fall faster than less massive objects? To answer the above questions, Newton's second law of motion (Fnet = m•a) will be applied to analyze the motion of objects that are falling under the sole influence of gravity (free fall) and under the dual influence of gravity and air resistance. Free Fall Motion As learned in an earlier unit, free fall is a special type of motion in which the only force acting upon an object is gravity. Objects that are said to be undergoing free fall, are not encountering a significant force of air resistance; they are falling under the sole influence of gravity. Under such conditions, all objects will fall with the same rate of acceleration, regardless of their mass . But why? Consider the free-falling motion of a 1000-kg baby elephant and a 1-kg overgrown mouse. If Newton's second law were applied to their falling motion, and if a free-body diagram were constructed, then it would be seen that the 1000-kg baby elephant would experiences a greater force of gravity.

This greater force of gravity would have a direct effect upon the elephant's acceleration; thus, based on force alone, it might be thought that the 1000-kg baby elephant would accelerate faster. But acceleration depends upon two factors: force and mass. The 1000-kg baby elephant obviously has more mass (or inertia). This increased mass has an inverse effect upon the elephant's acceleration. And thus, the direct effect of greater force on the 1000-kg elephant is offset by the inverse effect of the greater mass of the 1000-kg elephant; and so each object accelerates at the same rate - approximately 10 m/s/s. The ratio of force to mass (Fnet/m) is the same for the elephant and the mouse under situations involving free fall. This ratio (Fnet/m) is sometimes called the gravitational field strength and is expressed as 9.8 N/kg (for a location upon Earth's surface). The gravitational field strength is a property of the location within Earth's gravitational field and not a property of the baby elephant nor the mouse.

All objects placed upon Earth's surface will experience this amount of force (9.8 N) upon every 1 kilogram of mass within the object. Being a property of the location within Earth's gravitational field and not a property of the free falling object itself, all objects on Earth's surface will experience this amount of force per mass. As such, all objects free fall at the same rate regardless of their mass. Because the 9.8 N/kg gravitational field at Earth's surface causes a 9.8 m/s/s acceleration of any object placed there, we often call this ratio the acceleration of gravity. (Gravitational forces will be discussed in greater detail in a later unit of The Physics Classroom tutorial .) Falling with Air Resistance As an object falls through air, it usually encounters some degree of air resistance. Air resistance is the result of collisions of the object's leading surface with air molecules. The actual amount of air resistance encountered by the object is dependent upon a variety of factors. To keep the topic simple, it can be said that the two most common factors that have a direct effect upon the amount of air resistance are the speed of the object and the cross-sectional area of the object. Increased speeds result in an increased amount of air resistance.

Increased cross-sectional areas result in an increased amount of air resistance. Why does an object that encounters air resistance eventually reach a terminal velocity? To answer this questions, Newton's second law will be applied to the motion of a falling skydiver. In the diagrams below, free-body diagrams showing the forces acting upon an 85-kg skydiver (equipment included) are shown. For each case, use the diagrams to determine the net force and acceleration of the skydiver at each instant in time. Then use the button to view the answers. Physics Homework Help: Free Fall Problems Physics Homework Help Terminal Velocity / Other Problems Free fall refers to the motion of objects influenced only by gravity. The acceleration due to gravity, g, is different for different planets and at different heights above the planet. For motion near the Earth's surface (between sea level and the top of Mount Everest) g = 9.8 m/s2. (Related search term: free falling body problems; freefalling) QUICK QUESTION If you throw a rock downward from Mount Everest with negligible air resistance, then immediately after the rock leaves your hand the rock accelerates at a. less than 9.8 m/s2 b. 9.8 m/s2 c.

more than 9.8 m/s2 d. depends on the speed of the rock Answer to Quick Question The correct answer is b. Since there is negligible air resistance, any other answer confuses acceleration with speed. FREE FALL PROBLEM EXAMPLE 1. a. How long does it take a ball to fall from a roof to the ground 7.0 m below? b. With what speed does it strike the ground? ANSWER to FREE FALL PROBLEM EXAMPLE In kinematics problems, start off with a davfvit table. Use this format to list the information given and identify the quantity being solved for. Then identify the relationship between the given and the unknown quantities, substitute the values into the relationship, and solve fro the unknown. 1. frame of reference: down = + variable BACK TO TOP OF PAGE 1. A peso is dropped into a well and it falls for 3 seconds before hitting the water. What is the peso's average speed during its 3 second drop? 2. A rock is dropped from the top of an overhang and strikes the ground 6.5 seconds later. How high is the overhang in meters? 3.

It takes 0.210s for a dropped wrench to travel past a poster that is 1.35 meters tall. How high above the top of the poster was the wrench released? 4. A falling paratrooper (100 kg with parachute) experiences air resistance equal to 25% of his weight. What is his acceleration? 5. Consider a transparent elevator accelerating upward with acceleration equal to that of the gravity. If a rock was dropped inside the elevator, what would an observer on the ground see the rock do? THROW-UP EXAMPLES BACK TO TOP OF PAGE Throw up problems refer to situations where an object's initial velocity is opposite to its acceleration. The key is to choose a frame of reference. For example, if "up" is +, then "down" is -. The frame of reference must be used consistently throughout the solving process. THROW-UP PROBLEM EXAMPLE 2. How long does it take a ball to reach the ground 7.0 m below, if it is thrown straight up with an initial speed of 2.00 m/s? ANSWER to THROW-UP PROBLEM EXAMPLE 2. frame of reference: down = + variable 1.

You throw a ball downward from a window at a speed of 2.0 m/s. The ball accelerates at 9.8 m/s2. a. How fast is it moving when it hits the sidewalk 2.5 m below? b. If you throw the same ball up instead of down, how fast is it moving when it hits the sidewalk? 2. A ball is thrown straight up with a speed of 4.6 m/s. How long does the ball take to reach its maximum height? 3. A round is launched straight up at 460 m/s. How long will it take it to reach its apex and how high will that be? (air resistance may be neglected.) 4. What height will a dart achieve 7 seconds after being blown straight up at 50 m/s? 5. An apple thrown straight upward rises to 24 m above its launch point. At what height has apple's speed decreased to one-half of its initial value? 6. A stone is flung straight up from a point 1.50 m above the ground and with an initial speed of 19.6 m/s. a. What is the stone's maximum height above the ground? b.

How much time passes before the stone hits the ground? 7. A blimp is hovering above the ground. When the pilot drops a sandbag overboard, the blimp rises with a constant velocity of 2 m/s. At the moment the sandbag hits the ground, the blimp is 50 m above the ground. a. How far above the ground was the blimp when the sandbag was dropped? b. When the sandbag is halfway to the ground, what is its acceleration? 8. A helicopter is ascending vertically with a speed of 5.00 m/s. At a height of 105 m above the ground, a package is dropped from a window. How much time does it take for the package to reach the ground? 9. A rock is thrown up at an initial velocity of 9.8 m/s. What is the time it takes to hit the ground? CATCH-UP EXAMPLE BACK TO TOP OF PAGE In catch-up problems, two objects with different motions end up at the same place at the same time. Sometimes, these problems have the appearance of not having enough information to be solved. However, in physics we trust. These problems are complex because they describe two different motions.

The approach used is to simplify the problem by breaking it down into simple problems. This is down by using two columns in the davfvit table: one column for each motion. EXAMPLE CATCH-UP PROBLEM 3. A ball is dropped from a roof to the ground 8.0 m below. A rock is thrown down from the roof 0.600 s later. If they both hit the ground at the same time, what was the initial speed of the rock? ANSWER to EXAMPLE CATCH-UP PROBLEM 3. BACK TO TOP OF PAGE 1. An archer shoots an arrow with an initial velocity of 21 m/s straight up from his bow. He quickly reloads and shoots another arrow in the same way 3.0 s later. At what time and height do the arrows meet? 2. A hoist is lifting a naturalist to the top of a cliff at 2.03 m/s vertically. The naturalist suddenly realizes she has left her pet rock behind. A friend picks it up and tosses it straight upward. If the naturalist is 2.50 m above her friend, what is the minimum initial speed the pet rock must have to reach the naturalist?

3. A boy shoots a rock from his slingshot at a target just as the target drops from a tree branch. Should the boy aim a little above the target, a little below the target, or straight at the target? TERMINAL VELOCITY Terminal velocity is the maximum velocity an object can attain when it is subjected to a constant force (such as gravity) while travelling through a fluid (such as air). TERMINAL VELOCITY PROBLEM EXAMPLE 4. Describe the motion of a parachute. ANSWER to TERMINAL VELOCITY PROBLEM EXAMPLE 4. We presume the question is to describe the motion of an object attached to a parachute. Although parachutes have other uses, the discussion will be restricted to the motion of a falling payload, as opposed to the motion of a dragster, or aerobatic skydiver. In the most simplified case, the object continues to fall straight down. In a vacuum, or on a moon or planet with no atmosphere, a falling object will continue to accelerate until it hits the ground. However, in air (or any other fluid) drag is created as the object falls through the air. Drag is friction between the moving object and the surrounding air which resists the motion of the object.

Drag increases as the square of the velocity of the object, and also depends on the viscosity of the surrounding fluid. So when the speed of the object is 3 times greater, the drag on the object is 9 times greater. Also, there is more drag near the ground where the atmosphere is denser than at high altitudes where the air is “thinner.” As an object falls through the atmosphere picking up speed, drag becomes great enough that the object stops accelerating and continues falling at a constant speed. The maximum speed at which an object will fall is called “terminal velocity.” Drag also depends on the mass and surface area of the falling object. A feather takes longer to reach the ground than a much larger text book because of its ratio of surface area to mass. The larger the area and the less the mass, the lower the terminal velocity is. The main function of a parachute is to create drag, although some parachutes are designed to create lift. When the canopy opens, the effect is to increase the surface area of the falling object. The actual motion created by a parachute depends on the design of the parachute and the speed of the object when the parachute opens.

Parachutes can be designed to open slowly so that drag is increased slowly at a comfortable rate, or they can open more quickly which has a sudden pronounced effect and can cause strain or discomfort. One possibility is the object has not reached terminal velocity when the parachute opens. In this case the object will continue to pick up speed, but the downward acceleration of the falling object decreases to zero. At this point the object falls with constant speed. Another possibility is that the object has reached or exceeded terminal velocity with the parachute open. In this case the object will decelerate, continuing to fall at a slowing rate until terminal velocity is reached. At this point the object will fall with constant speed. D. Other Problems BACK TO TOP OF PAGE 1. What is drag? 2. What is the percent difference between the acceleration due to gravity at sea level and at the uppermost peak of Mount Everest? Answers to Free Fall Problems Selected solutions are printed below. For solutions to all the problems on this page click here. A3.

The average speed of the wrench as it passes the poster is 1.35 / 0.210 = 6.43 m/s Since acceleration is constant, the instantaneous speed of the wrench after falling 0.105s past the top of the poster is 6.43 m/s. Let the wrench have a speed, vt, at the top of the poster, an acceleration, a = 9.81 m/s/s, and a speed, vm = 6.43 m/s, at the time, t = 0.105s below the top of the poster. Applying a = (vf - vi)/t to the above scenario and substituting the values we obtain a = (vm - vt)/t 9.81 = (6.43 - vt)/(0.105) vt = 5.40 m/s When the wrench was initially dropped, it had a speed, vi = 0 and an acceleration, a = 9.81 m/s/s. As it reached the top of the poster a distance, d, below, it had a speed of 5.40 m/s.

Applying vf2 = vi2 + 2ad to the above scenario and substituting the values we obtain 5.402 = 02 + 2(9.81)d The Elephant and The Feather - with Air Resistance The Physics Classroom » Multimedia Studios » Newton's Laws » The Elephant and The Feather - with Air Resistance Elephant and Feather - Air Resistance Suppose that an elephant and a feather are dropped off a very tall building from the same height at the same time. We will assume the realistic situation that both feather and elephant encounter air resistance. Which object - the elephant or the feather - will hit the ground first? The animation at the right accurately depicts this situation. The motion of the elephant and the feather in the presence of air resistance is shown. Further, the acceleration of each object is represented by a vector arrow. Most people are not surprised by the fact that the elephant strikes the ground before the feather. But why does the elephant fall faster? This question is the source of much confusion (as well as a variety of misconceptions). Test your understanding by making an effort to identify the following statements as being either true or false. TRUE or FALSE: The elephant encounters a smaller force of air resistance than the feather and therefore falls faster.

The elephant has a greater acceleration of gravity than the feather and therefore falls faster. Both elephant and feather have the same force of gravity, yet the acceleration of gravity is greatest for the elephant. Both elephant and feather have the same force of gravity, yet the feather experiences a greater air resistance. Each object experiences the same amount of air resistance, yet the elephant experiences the greatest force of gravity. Each object experiences the same amount of air resistance, yet the feather experiences the greatest force of gravity. The feather weighs more than the elephant, and therefore will not accelerate as rapidly as the elephant. Both elephant and feather weigh the same amount, yet the greater mass of the feather leads to a smaller acceleration. The elephant experiences less air resistance and than the feather and thus reaches a larger terminal velocity. The feather experiences more air resistance than the elephant and thus reaches a smaller terminal velocity. The elephant and the feather encounter the same amount of air resistance, yet the elephant has a greater terminal velocity. If you answered TRUE to any of the above questions, then perhaps you have some confusion about either the concepts of weight, force of gravity, acceleration of gravity, air resistance and terminal velocity. The elephant and the feather are each being pulled downward due to the force of gravity.

When initially dropped, this force of gravity is an unbalanced force. Thus, both elephant and feather begin to accelerate (i.e., gain speed). As the elephant and the feather begin to gain speed, they encounter the upward force of air resistance. Air resistance is the result of an object plowing through a layer of air and colliding with air molecules. The more air molecules which an object collides with, the greater the air resistance force. Subsequently, the amount of air resistance is dependent upon the speed of the falling object and the surface area of the falling object. Based on surface area alone, it is safe to assume that (for the same speed) the elephant would encounter more air resistance than the feather. But why then does the elephant, which encounters more air resistance than the feather, fall faster? After all doesn't air resistance act to slow an object down? Wouldn't the object with greater air resistance fall slower? Answering these questions demands an understanding of Newton's first and second law and the concept of terminal velocity. According to Newton's laws, an object will accelerate if the forces acting upon it are unbalanced; and further, the amount of acceleration is directly proportional to the amount of net force (unbalanced force) acting upon it.

Falling objects initially accelerate (gain speed) because there is no force big enough to balance the downward force of gravity. Yet as an object gains speed, it encounters an increasing amount of upward air resistance force. In fact, objects will continue to accelerate (gain speed) until the air resistance force increases to a large enough value to balance the downward force of gravity. Since the elephant has more mass, it weighs more and experiences a greater downward force of gravity. The elephant will have to accelerate (gain speed) for a longer period of time before there is sufficient upward air resistance to balance the large downward force of gravity. Once the upward force of air resistance upon an object is large enough to balance the downward force of gravity, the object is said to have reached a terminal velocity. The terminal velocity is the final velocity of the object; the object will continue to fall to the ground with this terminal velocity. In the case of the elephant and the feather, the elephant has a much greater terminal velocity than the feather. As mentioned above, the elephant would have to accelerate for a longer period of time. The elephant requires a greater speed to accumulate sufficient upward air resistance force to balance the downward force of gravity.

In fact, the elephant never does reach a terminal velocity; the animation above shows that there is still an acceleration on the elephant the moment before striking the ground. If we were to depict the relative magnitude of the two forces acting upon the elephant and the feather at various times in their fall, perhaps it would appear as shown below. (NOTE: The magnitude of the force vector is indicated by the relative size of the arrow.) Observe from the above diagrams and the above animation that the feather quickly reaches a balance of forces and thus a zero acceleration (i.e., terminal velocity). On the other hand, the elephant never does reach a terminal velocity during its fall; the forces never do become completely balanced and so there is still an acceleration. If given enough time, perhaps the elephant would finally accelerate to high enough speeds to encounter a large enough upward air resistance force in order to achieve a terminal velocity. If it did reach a terminal velocity, then that velocity would be extremely large - much larger than the terminal velocity of the feather. So in conclusion, the elephant falls faster than the feather because it never reaches a terminal velocity; it continues to accelerate as it falls (accumulating more and more air resistance), approaching a terminal velocity yet never reaching it.

On the other hand, the feather quickly reaches a terminal velocity. Not requiring much air resistance before it ceases its acceleration, the feather obtains the state of terminal velocity in an early stage of its fall. The small terminal velocity of the feather means that the remainder of its fall will occur with a small terminal velocity. But what if there were no air resistance? If air resistance could somehow be eliminated (by conducting the experiment in a vacuum), then which object - the elephant or the feather - would strike the ground first? Investigate these questions by following the appropriate links to the Elephant and Feather (Free Fall) in the Multimedia Physics Studios. For more information on physical descriptions of motion, visit The Physics Classroom Tutorial . Detailed information is available there on the following topics: BBC - GCSE Bitesize: Falling objects Objects falling through a fluid eventually reach terminal velocity, when the resultant force acting on them is zero and they are moving at a steady speed. Weight is a force exerted on an object because of a gravitational force. Falling objects You should be able to describe the forces affecting a falling object at different stages of its fall. Usually, you need to think about two forces: The weight of the object.

This is a force acting downwards, caused by the object’s mass the Earth's gravitational field Air resistance. This is a frictional force acting in the opposite direction to the movement of the object. Three stages of falling When an object is dropped, we can identify three stages before it hits the ground: At the start, the object accelerates downwards because of its weight. There is no air resistance. There is a resultant force acting downwards. As it gains speed, the object's weight stays the same, but the air resistance on it increases. There is a resultant force acting downwards. Eventually, the object's weight is balanced by the air resistance. There is no resultant force and the object reaches a steady speed, called the terminal velocity. Check your understanding of this by studying the parachutist animation: Terminal velocity What happens if you drop a feather and a coin together? The feather and the coin have roughly the same surface area, so when they begin to fall they have about the same air resistance. As the feather falls, its air resistance increases until it soon balances the weight of the feather. The feather now falls at its terminal velocity. But the coin is much heavier, so it has to travel quite fast before air resistance is large enough to balance its weight.

In fact, it probably hits the ground before it reaches its terminal velocity. On the Moon An astronaut on the Moon carried out a famous experiment. He dropped a hammer and a feather at the same time and found that they landed together. The Moon's gravity is too weak for it to hold onto an atmosphere, so there is no air resistance. When the hammer and feather were dropped, they fell together with the same acceleration. Page:

Question: A falling object which ceases to accelerate due to the forces of drag and gravity being equal reaches what rate of descent (speed)?
Answer: Terminal Velocity